6
$\begingroup$

So given stationary colored gaussian noise $\mathbf{n}$, I know that I can decorrelate it by first finding it's autocorrelation $R_{nn}$ and performing $R^{-\frac{1}{2}}_{nn} \mathbf{n}$.

In practice of course, I need to estimate $R_{nn}$, which I can do via averaging $\mathbf{r}_{nn}$ or by fitting it to an AR model.

So, suppose I have a good estimate $\hat{R}_{nn}$, for a finite data record of length $N$. Such that the resulting spectrum of the noise frame is white enough, according to some metric.

Now, suppose I have a signal $\mathbf{x} = \mathbf{s} + \mathbf{n}$. I want to decorrelate the noise, such that after the decorrelation, I get a signal in white noise. From what I have read, the way to do this is to also apply $R^{-\frac{1}{2}}_{nn} \mathbf{x}$.

However, won't this distort the desired signal $\mathbf{s}$? Can anyone point me to some references regarding this?

Also: How else can I decorrelate colored noise in order to get my signal embedded in white noise?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Yes, the "Whitening" is basically filtering the signal using an LTI System.

The Result would be the signal $ \mathbf{s} $ filtered by the system which whitens the noise.

In the framework of Matched Filter this is OK since you will be able to filter the Reference Signal as well and hence have the same performance.

Yet for other usages you must take under consideration the effect of whitening system.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I figured. The autocorrelation of colored noise decays exponentially. Hence, if we were to whiten it, we would need a function that increases exponentially. Therefore, it seems to me that a whitening filter will act to boost the high frequencies of the signal, while nulling the lower frequencies. $\endgroup$
    – The Dude
    May 16, 2016 at 18:15
  • $\begingroup$ Well, that depends how the Function which colored the noise looked like. If it was "High Pass Filter" the result would be exactly the opposite of what you described. $\endgroup$
    – Royi
    May 16, 2016 at 19:09
  • $\begingroup$ 5 years later and you finally get it LOL $\endgroup$
    – The Dude
    Mar 26, 2021 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.